Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int n;
int a[10100];
int b[10100];
int ans[1001000][2];
int ansl;
int c[10100];
int d[500100][2];
int dl;
void gelim() {
  int i, j, p;
  p = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < p; j++) {
      if ((c[i] ^ c[j]) < c[i]) {
        c[i] ^= c[j];
        d[dl][0] = i;
        d[dl][1] = j;
        dl++;
      }
    }
    if (c[i] != 0) {
      if (i == p)
        p++;
      else {
        c[p] ^= c[i];
        d[dl][0] = p;
        d[dl][1] = i;
        dl++;
        c[i] ^= c[p];
        d[dl][0] = i;
        d[dl][1] = p;
        dl++;
        p++;
      }
      j = p - 1;
      while (j != 0 && c[j - 1] < c[j]) {
        c[j] ^= c[j - 1];
        d[dl][0] = j;
        d[dl][1] = j - 1;
        dl++;
        c[j - 1] ^= c[j];
        d[dl][0] = j - 1;
        d[dl][1] = j;
        dl++;
        c[j] ^= c[j - 1];
        d[dl][0] = j;
        d[dl][1] = j - 1;
        dl++;
        j--;
      }
    }
  }
}
int main() {
  int i, j;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &b[i]);
  }
  for (i = 0; i < n; i++) {
    c[i] = a[i];
  }
  gelim();
  for (i = 0; i < dl; i++) {
    ans[ansl][0] = d[i][0];
    ans[ansl][1] = d[i][1];
    ansl++;
  }
  dl = 0;
  for (i = 0; i < n; i++) {
    a[i] = c[i];
  }
  for (i = 0; i < n; i++) {
    c[i] = b[i];
  }
  gelim();
  for (i = 0; i < n && a[i] != 0; i++) {
    for (j = i; j < n && a[j] != 0; j++) {
      if ((a[i] ^ c[i]) > ((a[i] ^ c[i]) ^ a[j])) {
        a[i] ^= a[j];
        ans[ansl][0] = i;
        ans[ansl][1] = j;
        ansl++;
      }
    }
    if (a[i] != c[i]) {
      printf("-1");
      return 0;
    }
  }
  if (c[i + 1] != 0) {
    printf("-1");
    return 0;
  }
  for (i = dl - 1; i >= 0; i--) {
    ans[ansl][0] = d[i][0];
    ans[ansl][1] = d[i][1];
    ansl++;
  }
  printf("%d\n", ansl);
  for (i = 0; i < ansl; i++) {
    printf("%d %d\n", ans[i][0] + 1, ans[i][1] + 1);
  }
}